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copied!<p>I've used this and it works quite well. <a href="http://en.wikipedia.org/wiki/Newton's_method#Examples">Here's an example of how the algorithm works at a high level.</a></p> <p>Edit: I was curious to see just how accurate this was as defined below. Here is the sqrt(2) from an <a href="http://apod.nasa.gov/htmltest/gifcity/sqrt2.1mil">official source</a>:</p> <pre><code>(first 200 digits) 1.41421356237309504880168872420969807856967187537694807317667973799073247846210703885038753432764157273501384623091229702492483605585073721264412149709993583141322266592750559275579995050115278206057147 </code></pre> <p>and here it is using the approach I outline below with <code>SQRT_DIG</code> equal to 150:</p> <pre><code>(first 200 digits) 1.41421356237309504880168872420969807856967187537694807317667973799073247846210703885038753432764157273501384623091229702492483605585073721264412149709993583141322266592750559275579995050115278206086685 </code></pre> <p>The first deviation occurs after <strong>195 digits of precision</strong>. Use at your own risk if you need such a high level of precision as this.</p> <p>Changing <code>SQRT_DIG</code> to 1000 yielded <strong>1570 digits of precision</strong>.</p> <pre><code>private static final BigDecimal SQRT_DIG = new BigDecimal(150); private static final BigDecimal SQRT_PRE = new BigDecimal(10).pow(SQRT_DIG.intValue()); /** * Private utility method used to compute the square root of a BigDecimal. * * @author Luciano Culacciatti * @url http://www.codeproject.com/Tips/257031/Implementing-SqrtRoot-in-BigDecimal */ private static BigDecimal sqrtNewtonRaphson (BigDecimal c, BigDecimal xn, BigDecimal precision){ BigDecimal fx = xn.pow(2).add(c.negate()); BigDecimal fpx = xn.multiply(new BigDecimal(2)); BigDecimal xn1 = fx.divide(fpx,2*SQRT_DIG.intValue(),RoundingMode.HALF_DOWN); xn1 = xn.add(xn1.negate()); BigDecimal currentSquare = xn1.pow(2); BigDecimal currentPrecision = currentSquare.subtract(c); currentPrecision = currentPrecision.abs(); if (currentPrecision.compareTo(precision) <= -1){ return xn1; } return sqrtNewtonRaphson(c, xn1, precision); } /** * Uses Newton Raphson to compute the square root of a BigDecimal. * * @author Luciano Culacciatti * @url http://www.codeproject.com/Tips/257031/Implementing-SqrtRoot-in-BigDecimal */ public static BigDecimal bigSqrt(BigDecimal c){ return sqrtNewtonRaphson(c,new BigDecimal(1),new BigDecimal(1).divide(SQRT_PRE)); } </code></pre> <p>be sure to check out barwnikk's answer. it's more concise and seemingly offers as good or better precision.</p>

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